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 Question from Sandra, a student: Show that x^3 - 3kx + 1 = 0 does not have two distinct roots when k < 0.

Hi Sandra,

In fact if k < 0 then x3 - 3kx + 1 = 0 only has one root.

x3 - 3kx + 1 = 0 is the same as

x(x2 - 3k) = -1

Now if k is negative, then x2 - 3k must be positive. So x times some positive number equals -1. Thus x is negative. Thus if k < 0 then every root of x3 - 3kx + 1 = 0 is negative.

Suppose there are two roots a and b, then (x - a) and (x - b) are factors of x3 - 3kx + 1 and hence

x3 - 3kx + 1 = (x - a)( x - b) g(x)

for some polynomial g(x). Since the original polynomial is a cubic, g(x) is linear and since the leading coefficient of the cubic is 1, g(x) = x - c for some real number c. Hence

x3 - 3kx + 1 = (x - a)( x - b)(x - c)

Thus a, b and c are roots of the cubic so they are all negative.

What do you know about a b c?

Stephen and Penny

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